Year	Quarters
	I	II	III	IV
1994	78	66	84	80
1995	76	74	82	78
1996	72	68	80	70
1997	74	70	84	74
1998	76	74	86	82
Total	376	352	416	384
Average	75.2	70.4	83.2	76.8

Grand average = \(\frac {75.2+70.4+83.2+76.8}{4}\)
= \(\frac {305.6}{4}\) = 76.4
Seasonal index S.I = \(\frac {Quarterly average}{Grand average} \times 100\)
Hence, S.I for I quarter = \(\frac {75.2}{76.4} \times 100\) = 98.4
S.I for II quarter =  \(\frac {70.4}{76.4} \times 100\) = 92.14
S.I for III quarter =  \(\frac {83.2}{76.4} \times 100\) = 108.9
S.I for IV quarter =  \(\frac {76.8}{76.4} \times 100\) = 100.5

x	y	x2	xy
0	1	0	0
1	1	1	1
2	3	4	6
3	4	9	12
4	6	16	24
10	15	30	43

Let y= ax + b be the best line of fit.
The normal equations are
a\(\sum\)x + nb = \(\sum\)y
a\(\sum\)x² + b\(\sum\)x = \(\sum\)xy
10a+ 5b = 15
⇒ 2a+b = 3 ...(1)
30a + 10b = 43
⇒  3a+b =  4.3 ....(2)
(1)-(2) ⇒ - a = -1.3
⇒  a = 1.3
Substituting a = 1.3 in (1) we get,
2(1.3) + b = 3
⇒ 2.6 + b = 3
⇒ b = 3 - 2.6 = 0.4
∴ The line of best fit isy = 1.3x + 0.4

 

 

Given sample size n = 50
Sample mean \(\bar { x }\)  = 825
Population mean μ = 900
Population S.D. σ = 110
Null hypotheses: H₀: μ = 900
Alternative hypotheses: H₁: μ ≠ 900
The test statistic, z = \(\frac { \bar { x } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)
=\(\frac { 825-900 }{ \frac { 110 }{ \sqrt { 50 } } } \) = -4.82
∴ |z| = -4.82
As the significance level is α = 0.05, \(Z_{ \frac { \alpha }{ 2 } }\) = 1.96
Here |z| > \(Z_{ \frac { \alpha }{ 2 } }\) as 4.82 > 1.96
Inference: As |z| > \(Z_{ \frac { \alpha }{ 2 } }\), H₀ is rejected. Hence, we can conclude that mean life time of the population of electric bulbs cannot be taken as 900 hours.

Sample size n = 50
Sample mean \(\bar { x } \) = 75
Sample S.D. s = 10
Standard error (S.E) = \(\frac { s }{ \sqrt { n } } =\frac { 10 }{ \sqrt { 50 } } =\frac { 10 }{ 7.07 } \)
= 1.414
As the significance level is α = 0.005, \(Z_{ \frac { \alpha }{ 2 } }\) = 1.96
∴ 95% confidence limits for the population mean is \(\bar { X } -Z_{ \frac { \alpha }{ 2 } }(S.E)\le \bar { X } -Z_{ \frac { \alpha }{ 2 } }(S.E)\)
⇒ 75 - (1.96)(1.414) ≤ μ ≤ 75 + (1.96)(1.414)
⇒ 75-2.771 ≤ μ ≤ 75 + 2.771
⇒ 72.23 ≤ μ ≤ 77.77
Hence, the 95% confidence interval of the population mean is (72.23, 77.77)

Let X denote the life of army shoes.
Given μ = 8, σ = 2 and N = 5000
When μ = 12, Z = \(\frac { X-\mu }{ \sigma } =\frac { 12-8 }{ 2 } \) = 2
∴ P(X≤12) = P(Z≤2)
= P(-∞ = P(-∞ =0.5+0.4772
P(X≤12) = 0.9772

∴ The probability for a shoe need to be replaces is 0.9772.
∴ Out of 5000 pairs of shoes, number of pairs need to be replaced = 5000 \(\times\) 0.9772
 = 4886.

The p.d.f for the normal distribution is
f(x) = \(\frac { 1 }{ \sigma \sqrt { 2\pi } } .e^{ -\frac { 1 }{ 2 } \left( \frac { x-\mu }{ \sigma } \right) ^{ 2 } }\), -∞ Given
f(x) = \(\frac { 1 }{ K } { e }^{ 8x-4x^{ 2 } }\)
= \(\frac { 1 }{ K } e^{ -4 }({ x }^{ 2 }-2x)\)
= \(\frac { 1 }{ K } e^{ -4 }({ x }^{ 2 }-2x+1-1)\)
\(\frac { 1 }{ K } e^{ -4 }(x-1)^{ 2 }+4\)
=\(\frac { 1 }{ K } .{ e }^{ 4 }e^{ -\frac { 1 }{ 2 } \frac { x-1^{ 2 } }{ \frac { 1 }{ 8 } } }\)
=\(\frac { 1 }{ 4 } { e }^{ 4 }e^{ -\frac { 1 }{ 2 } \left( \frac { x-1 }{ \frac { 1 }{ \sqrt { 8 } } } \right) ^{ 2 } }\) ...(2)
From (1) & (2), μ = 1, σ = \(\frac { 1 }{ \sqrt { 8 } } \) and
\(\frac { 1 }{ \sigma \sqrt { 2\pi } } =\frac { 1 }{ K } e^{ 4 }\)
\(\frac { 1 }{ \frac { 1 }{ \sqrt { 8 } } .\sqrt { 2 } \pi } =\frac { 1 }{ K } e^{ 4 }\)
⇒ \(\frac { \sqrt { 8 } }{ \sqrt { 2\pi } } =\frac { 1 }{ K } e^{ 4 }\)
\(\sqrt { \frac { 4 }{ \pi } } =\frac { 1 }{ K } { e }^{ 4 }\)
∴ K = \({ e }^{ 4 }\times \sqrt { \frac { \pi }{ 4 } } \)

The probability distribution of X is

X	0	1	2	3	4
P(X)	0.1	k	2k	2k	k

Σp_i = 1 ⇒ 0.1 + k + 2k + 2k + k = 1
⇒ 0.1 + 6k = 1 ⇒ 6k = 1-0.1 = 0.9
⇒ k = \(\frac{0.9}{6}\) = 0.15
And the probability that you study atleast 2 hours is P(X≥2)=P(X=2)+P(X=3)+P(X=4)
= 2k + 2k + k = 5k
= 5(0.15) = 0.75

Since the random variable X is the probability distribution function, Σp_i = 1
∴ a + 2a + 3a + 4a + 5a + 6a = 1
21a = 1 ⇒ a = \(\frac{1}{21}\)
Now, F(4) = P(X ≤ 4)
= P(X = 0) + P(X = 1) + P(X = 2)P(X = 3) + P(X = 4)
= a + 2a + 3a + 4a + 5a = 15a
= 15\((\frac{1}{21})=\frac{5}{7}\)
∴ F(4) = \(\frac{5}{7}\)

Here the number of rows and columns are equal.
∴ The given assignment problem is balanced.
Step 1 :
Select a minimum element in each row and subtract this from all the elements in its row.

Here IV column has no zero. Go to step 2.
Step 2:
Select the minimum element in each column and subtract this from all the elements in its column.

Since each row and column contains atleast one zero, assignments can be made.
Step 3:
Examine the rows with only one zero. Mark that. zero by ׈ and draw a vertical line.

Thus, all the assignments have been made.
The optimal assignment schedule and total cost is
 

Job	Operator	Cost
A	III	2
B	IV	6
C	II	4
D	I	5
Total Cost		Rs. 17

Σa_i = 12 + 14 + 4 = 30
Σb_j = 9 + 10 + 11 = 30
Σa_i =  Σb_j
∴ The given problem is a balanced transportation problem.
Hence, there exists a feasible solution to the given problem.
I - allocation :

[∵ the highest penalty is 7, In C, least cost is 0 & min (11, 14) = 11] 
II - allocation :

[∵ the highest penalty is 4, In S1'least cost is 1& min (10, 12) = 10]
III - allocation :

[∵ In A, least cost is 2 & min (9, 3) = 3]
IV - allocation :

[∵ In A, least cost is 3 & min (6,4) = 4]
V - allocation :

[∵ min (2, 2) = 2]
Thus, the allocations are

∴ The transportation schedule is
S₁ → A, S₁ → B, S₂ → A, S₂ → C  S₃ → A
Hence, the total transportation cost is
= 2(5) + 10(1) + 3(2) + 11(0) + 4(3)
= 10 + 10 + 6 + 0 + 12 = Rs. 38

\(\int { \frac { { ({ a }^{ x }{ +b }^{ x }) }^{ 2 } }{ { a }^{ x }b^{ x } } dx } \) = \(\int { \frac { { a }^{ 2x }+{ b }^{ 2x }+{ 2a }^{ x }{ b }^{ x } }{ { 2a }^{ x }{ b }^{ x } } } \)
[∵ (a + b)² = a² + 2ab + b²]
= \(\int { \left( \frac { { a }^{ 2 }x }{ { a }^{ x }{ b }^{ x } } +\frac { { b }^{ 2x } }{ { a }^{ x }{ b }^{ x } } +\frac { 2{ a }^{ x }{ b }^{ x } }{ { a }^{ x }{ b }^{ x } } \right) } dx\)
= \(\int { \left( \frac { { a }^{ x } }{ { b }^{ x } } +\frac { { b }^{ x } }{ { a }^{ x } } +2 \right) dx } \)
= \(\int { \left( { \left( \frac { a }{ b } \right) }^{ x }+{ \left( \frac { b }{ a } \right) }^{ x }+2 \right) } dx\)
= \(\frac { { \left( \frac { a }{ b } \right) }^{ x } }{ { log }_{ e }\left( \frac { a }{ b } \right) } +\frac { { \left( \frac { b }{ a } \right) }^{ 2 } }{ { log }_{ e }\left( \frac { b }{ a } \right) } +2x+c\)

\(\int { \frac { cos2x-cos2\alpha }{ cosx-cos\alpha } } dx\)
= \(\frac { ({ 2cos }^{ 2 }x-1)-({ 2cos }^{ 2 }\alpha -1) }{ cos\quad x-cos\alpha } dx\)
= \(\frac { { 2cos }^{ 2 }x-cos^{ 2 }\alpha }{ cos\quad x-cos\alpha } \)

= \(2\int { (cos\quad x+cos\alpha )dx } \)
= \(2[sinx+cos\alpha .x]+c\)
= \(2sinx+2xcos\alpha +c\)

The auxiliary equation is m² + 1 = 0
⇒  m² = -1
⇒ m = ±\(\sqrt { -1 } \) = ±i
Here α = 0, β = 1
∴ CF is e^0x [A cosx + B sinx]
∴ The general solution is
y = A cos x + B sin x ...(1)
Given when x = 0, y = 2
∴ 2 = A cos 0 + B sin 0
⇒ 2 = A+0 ⇒ A = 2
[∵ cos0 = 1 and sin0 = 0]
Also, when x = \(\frac { \pi }{ 2 } \), y = -2
∴ -2 = A\(cos\frac { \pi }{ 2 } +Bsin\frac { \pi }{ 2 } \)
⇒ -2 = A(0)(+B(1) ⇒ B = -2
[∵ \(cos\frac { \pi }{ 2 } \) = 0 and \(sin\frac { \pi }{ 2 } \)= 1]
Substituting the values of A & B in (1) we get,
y = 2 cos x - 2 sin x
⇒ y = 2 (cosx - sin x)

Given (x²-yx²)dy + (y²+xy²)dx = 0
⇒ x²(1-y)dy+y²(1+x)dx = 0
⇒ x²(1-y)dy = -y²(1+x)dx
Separating the variables we get,
\(\frac { (1-y) }{ y^{ 2 } } dy=-\frac { (1+x) }{ x^{ 2 } } \)dx
⇒ \(\frac { 1 }{ { y }^{ 2 } } dy-\frac { 1 }{ y } dy=-\frac { 1 }{ x^{ 2 } } dx-\frac { 1 }{ x } dx\)
Integrating, \(\int { { y }^{ -2 } } dy-\int { \frac { 1 }{ y } } dy=-\int { \frac { 1 }{ x^{ 2 } } } dx-\int { \frac { 1 }{ x } } \)
\(-\frac { 1 }{ y } -logy=\frac { 1 }{ x } \) -log x + C
⇒ log x - log y = \(\frac { 1 }{ x } +\frac { 1 }{ y } \)+C
⇒ log \(log\left( \frac { x }{ y } \right) =\frac { x+y }{ xy } \)+C
⇒ \(\frac { x }{ y } =e^{ \frac { x+y }{ xy } +C }\)
⇒ \(\frac { x }{ y } =K.e^{ \frac { x+y }{ xy } }\)  [where e^C = K]

Let us calculate the number of men whose wages is less than Rs. 35 by using Newton's forward interpolation formula
x_o + nh = x ⇒ 30 + n (10) = 35
⇒ 10n = 35 - 30 = 5
⇒ n = \(\frac{5}{10}\) = 0.5
The difference table is

\(\Rightarrow { y }_{ o }+\frac { n }{ n! } \triangle { y }_{ o }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ o }+\frac { n(n+1)(n-2) }{ 3! } { \triangle }^{ 3 }{ (y }_{ o })\)
\(y(35)=9+\frac { 0.5 }{ 1! } (30)+\frac { (0.5)(0.5-1) }{ 2 } (5)+\frac { (0.5)(0.5-1)(0.5-2) }{ 6 } (2)\)
= 9 + 15 - 0.6 + 0.1 = 24 (approximately)
∴ Number of men getting wages between Rs. 30 and Rs. 35 is y(35) - y(30) = 24 - 9 = 15.

From the graph, it is clear that when x = 27, the value of y is 24.8

Given \(R'(x)=\frac { 3 }{ { x }^{ 2 } } -\frac { 2 }{ x } \)
\(\Rightarrow \int { R'(x) } =\int { \left( \frac { 3 }{ { x }^{ 2 } } -\frac { 2 }{ x } \right) } dx\)
\(\Rightarrow R(x)=\frac { -3 }{ x } -2log\quad x+k\)
Given R(1) = 6 ⇒ when x = 1, R = 6
\(\Rightarrow 6=\frac { -3 }{ 1 } -2log1+k\)
⇒ 6 + 3 = k [∵ log 1 = 0]
⇒ k = 9 
\(\therefore R(x)=-\frac { 3 }{ x } -2log\quad x+9\)
Demand function \(P=\frac { R }{ x } \)
\(\\ =\frac { 3 }{ { x }^{ 2 } } -\frac { 2log\quad x }{ x } +\frac { 9 }{ x } \)

Given, marginal cost, C' (x) \(\frac{x}{3000}+2.50\)
\(\int { C'(x) } =\int { \left( \frac { x }{ 300 } +2.50 \right) dx } \)
\(C(x)=\frac { { x }^{ 2 } }{ 6000 } +2.50x+k\)
When x = 0, c = 0 ⇒ k = 0
∴ c(x) = \(\frac{x^2}{6000}+2.50x\)
When x = 3000
Cost of production
\(=\frac { { (3000) }^{ 2 } }{ 6000 } +2.50(3000)\)

= 1500 + 7500
= Rs. 9000

Given non-homogeneous equations are
2x + 3y = 5  6X + 5y = 11
\(\Delta =\left| \begin{matrix} 2 & 3 \\ 6 & 5 \end{matrix} \right| =10-18=-8\)
Since \(\Delta \neq 0\) Cramer's rule can be applied and the system is consistent with unique solution
\(\Delta x=\left| \begin{matrix} 5 & 3 \\ 11 & 5 \end{matrix} \right| =25-33=-8\)
\(\Delta y=\left| \begin{matrix} 2 & 5 \\ 6 & 11 \end{matrix} \right| =22-30=-8\)
\(\therefore x=\cfrac { \Delta x }{ \Delta } =\cfrac { -8 }{ -8 } =1\)
\(y=\cfrac { \Delta y }{ \Delta } =\cfrac { -8 }{ -8 } =1\)
\(\therefore \) Solution set is {1, 1}

\(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 2 \\ -1 \\ 3 \end{matrix} \right] \)
\(\Rightarrow \left( \begin{matrix} x0+0 \\ 0+0+z \\ 0+y+0 \end{matrix} \right) =\left( \begin{matrix} 2 \\ -1 \\ 3 \end{matrix} \right) \)
\(\Rightarrow \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 2 \\ -1 \\ 3 \end{matrix} \right) \)
\(\Rightarrow x=2\quad z=-1\quad y=3\)
\(\therefore\) Solution set is {2,3, -1}

A =\(\left( \begin{matrix} 3 & 1 & -5 \\ 1 & -2 & 1 \\ 1 & 5 & -7 \end{matrix}\begin{matrix} -1 \\ -5 \\ 2 \end{matrix} \right) \)
The order of A is 3 x 4
\(\therefore \rho (A)\le \text{minimum} \ of(3,4)\)
\(\therefore \rho (A)\le 3\)
Let us transform the matrix A to an echelon form

Matrix A	Elementary Transformation
\(A=\left( \begin{matrix} 3 & 1 & -5 \\ 1 & -2 & 1 \\ 1 & 5 & -7 \end{matrix}\begin{matrix} -1 \\ -5 \\ 2 \end{matrix} \right) \)
\(\sim \left( \begin{matrix} 1 & 5 & -7 \\ 1 & -2 & 1 \\ 3 & 1 & -5 \end{matrix}\begin{matrix} 2 \\ -5 \\ -1 \end{matrix} \right) \)	\({ R }_{ 1 }\leftrightarrow { R }_{ 3 }\)
\(\sim \left( \begin{matrix} 1 & 5 & -7 \\ 0 & -7 & 8 \\ 3 & 1 & -5 \end{matrix}\begin{matrix} 2 \\ -7 \\ -1 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }+{ R }_{ 1 }\)
\(\sim \left( \begin{matrix} 1 & 5 & -7 \\ 0 & -7 & 8 \\ 0 & -14 & 16 \end{matrix}\begin{matrix} 2 \\ -7 \\ -7 \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }+{ 3R }_{ 1 }\)
\(\sim \left( \begin{matrix} 1 & 5 & -7 \\ 0 & -7 & 8 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} 2 \\ -7 \\ 7 \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ 2R }^{ 2 }\)

The matrix is in echelon form and the number of non-zero matrix is 3
\(\therefore \rho (A)=3\)

Scale:
In x axis 1 cm = 10 units
In y axis 1 cm = 2 units
Plot the points (40, 6.2), (50, 7.2), (60, 9.1) and (70, 12). At x = 48, draw a vertical line to the graph and from the intersecting point, draw a horizontal line to meet the y-axis
From the graph, we find that when x = 48, the value of Y is equal to 6.8.

	Estimated Conditional Profit 0
Rainfall	High	Medium	Low	Minimum payoff	Maximum payoff
Crop A	8000	4500	2000	2000	8000
Crop B	3500	4500	5000	3500	5000
Crop C	5000	5000	4000	4000	5000

(i) Max (2000,3500,4000) = 4000
∴ Crop C is the best according to maximin criteria
(ii) Min (8000,5000,5000) = 5000
∴ Crop B and C are best according to minimax criteria

(i) Take a suitable scale for the values of x and y, and plot the various points on the graph paper for given values of x and y.
(ii) Draw a suitable curve passing through the plotted points.
(iii) Find the point corresponding to the value x = 38 on the curve and then read the corresponding value of y on the y- axis, which will be the required interpolated value.
From the graph in Figure we find that for x = 38, the value of y is equal to 35

(D²−4D−1)y = e^−3x
The auxiliary equation is
m²−4m−1 = 0
(m−2)²−4−1 = 0
(m− 2)² = 5
\(m-2=±\sqrt { 5 } \)
\(m=2±\sqrt5\)
C.F = Ae^{\((2+\sqrt5)x\)} + Be^{\((2-\sqrt5)x\)}
\(PI=\frac { 1 }{ \phi (D) } f(x)\)
= \(\frac { 1 }{ { D }^{ 2 }-4D-1 } { e }^{ -3x }\)
\(=\frac { 1 }{ { (-3) }^{ 2 }-4(-3)-1 } { e }^{ -3x }\) (Replace D by −3)
\(=\frac { 1 }{ 9+12-1 } ​​{ e }^{ -3x }\)
\(=\frac { e^{ -3x } }{ 20 } \)
Hence the general solution is y = C.F+P.I
⇒ y = Ae^{\((2+\sqrt5)x\)} + Be^{\((2-\sqrt5)x\)}+\(\frac { e^{ -3x } }{ 20 } \)

Here total supply = 5 + 8 + 7 + 14 = 34, Total demand = 7 + 9 + 18 = 34
(i.e) Total supply =Total demand 
Therefore The given problem is balanced transportation problem.
\(\therefore\) we can findan initial basic feasible solution to the given problem.
From the above table we can choose the cell in the North West Corner. Here the cell is (1, A)
Allocate as much as possible in this cell so that either the capacity of first row is exhausted or the destination requirement of the first column is exhausted.
i.e. x₁₁ = min (5, 7) = 5

Reduced transportation table is

Now the cell in the North west corner is (2,  A)
Allocate as much as possible in the first cell so that either the capacity of second row is exhausted or the destination requirement of the first column is exhausted.
i.e. x₁₂ = min (2, 8) = 2

Reduced transportation table is

Here north west corner cell is (2, B) Allocate as much as possible in the first cell so that either the capacity of second row is exhausted or the destination requirement of the second column is exhausted.
i.e. x₂₂ = min (6, 9) = 6

Reduced transportation table is

Here north west corner cell is (3,B).
Allocate as much as possible in the first cell so that either the capacity of third row is exhausted or the destination requirement of the second column is exhausted.
i.e. x₃₂ = min (7, 3) = 3

Reduced transportation table is

Here north west corner cell is (3,C) Allocate as much as possible in the first cell so that either the capacity of third row is exhausted or the destination requirement of the third column is exhausted.
i.e. x₃₃ = min (4, 18) = 4

Reduced transportation table and final allocation is x₄₄ = 14

Thus we have the following allocations

Transportation schedule : 1⟶A, 2⟶B, 3⟶B, 3⟶C, 4⟶C
The total transportation cost.
= (5 \(\times\) 2) + (2 x\(\times\) 3) + (6 \(\times\) 3) + (3 \(\times\) 4) + (4 \(\times\) 7) + (14 \(\times\) 2)
= Rs. 102

The line of least fit is a line from which the sum of the deviations of various points is zero. This is the best method for obtaining the trend values.
(i) The straight line trend is represented by
 Y = a + b X ...(1)
where Y is the actual value and X is time
(ii)  The constants 'a' and 'b' are estimated by solving the following two normal Equations
 \(\sum\) Y = n a + b \(\sum\) X
\(\sum\) XY = a\(\sum\) X + b  \(\sum\) X²   where n is the number of years given in the data.
(iii) By substituting the values of 'a' and 'b' in the trend equation (1), we get the line of best fit.

Tendency movements are due to nature, which repeat themselves periodically in every seasons. These variations repeat themselves in less than one year time. It is measured in an interval of time.
Seasonal variations may be influenced by natural force, social customs and traditions.

Sample size n = 100,
Sample mean \(\\ \bar { X } =3.5\)
Population mean μ = 4
Population standard deviation σ = 3
Null Hypotheses: There is no significant difference in the mean. i.e., H_o : μ = 4
Alternative Hypotheses : There is Significant difference in the mean.
i.e., H₁ : μ ≠ H
The level of significance ∝ = 5% = 0.05
Applying the test statistic,\(Z=\frac { \bar { X } -\mu }{ \frac { \sigma }{ \sqrt { n } } } \)
\(\Rightarrow Z=\frac { 3.5-4 }{ \frac { 3 }{ \sqrt { 100 } } } =\frac { -.5 }{ .3 } =-1.667\)
\(\Rightarrow |Z|=1.667\)
\({ Z }_{ \frac { \alpha }{ 2 } }=1.96\)
Here Z < \({ Z }_{ \frac { \alpha }{ 2 } }\)i.e., 1.667<1.96
Inference: Since Z<\({ Z }_{ \frac { \alpha }{ 2 } }\)at 5% level of significance, the null hypothesis H₀ is accepted. Hence there is no Significant difference in the mean.

Sample size n = 50
Sample S.D s = 6.3
Population S.D \(\sigma\) = 6
The standard error for sample S.D is given by 
\(S.E=\sqrt { \frac { { \sigma }^{ 2 } }{ 2n } } =\frac { 6 }{ \sqrt { 2(50) } } =\frac { 6 }{ \sqrt { 100 } } =0.6\)
Thus standard error for sample S.D = 0.6.

\(Let\ I=\int { \frac { { e }^{ x }dx }{ { e }^{ 2x }-9 } } dx\)
\(=\int { \frac { { e }^{ x }dx }{ { \left( { e }^{ x } \right) }^{ 2 }-9 } } \)
\(Put\ { e }^{ x }=t\Rightarrow { e }^{ x }dx=dt\)
\(\therefore I=\int { \frac { dt }{ { t }^{ 2 }-9 } } =\int { \frac { dt }{ { t }^{ 2 }-{ 3 }^{ 2 } } } \)
\(=\frac { 1 }{ 2\times 3 } \log { \left| \frac { t-3 }{ t+3 } \right| } +c\)
\(\left[ \because \int { \frac { dx }{ { x }^{ 2 }-{ a }^{ 2 } } } =\frac { 1 }{ 2a } \log { \left| \frac { x-a }{ x+a } \right| } +c \right] \)
\(=\frac { 1 }{ 6 } \log { \left| \frac { { e }^{ x }-3 }{ { e }^{ x }+3 } \right| } +c\)

Probability of bus running late is denoted as p = 1-0.8 = 0.2
Probability of bus running according to the schedule is q = 0.8
Also given that n = 10
The binomial distribution is p(x) = 10C_x(0.2)^x(0.8)^10-x
(i) probability that exactly one is late P(x = 1) = 10C₁pq⁹
= 10C₁(0.2)(0.8)⁹
(ii) probability that at least one is late
= 1 – probability that none is late
= 1 – p(x = 0)
= 1– (0.8)¹⁰

The parameters of Binomial distribution are n and p
For Binomial distribution Mean = np = 20
Standard deviation =\(\\ \\ \\ \sqrt { npq } =4\)   
∴ npq = 16
⇒ npq/np = 16/20 = 4/5
q = 4/5
⇒ p = 1–q = 1– (4/5) = 1/5
Since np = 20
n = \(\frac{20}{p}\)
n = 100

Given probability distribution function is 
X=x 1 3 5 P(X=x) 2k 3k 4k a) Since the given function is a p.d.f. each p_i≥0 and Σp_i=1
⇒2k + 3k+4k = 1 ⇒ 9k = 1 ⇒ k = \(\frac{1}{9}\)
b) P(X>2) = P(X = 3) + P(X = 5)
= 3k + 4k = 7k = 7(\(\frac{1}{9}\))
\(\left[ \because k=\frac { 1 }{ 9 } \right] \)
\(\therefore P(X\ge 2)=\frac { 7 }{ 9 } \)

Given p.d.f is
\(f(x)= \begin{cases}\frac{1}{30} e^{-\frac{x}{30}}, & \text { for } x>0 \\ 0, & \text { for } x \leq 0\end{cases}\)
Expected number of miles
\(E(X)=\int _{ -\infty }^{ \infty }{ x.f(x)dx } \)
\(=\int _{ 0 }^{ \infty }{ x.\frac { 1 }{ 30 } { e }^{ \frac { -x }{ 30 } }dx=\int _{ }^{ \infty }{ { xe }^{ \frac { -x }{ 36 } }dx } } \)
\(\left[ \because \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -axdx }=\frac { n! }{ { a }^{ n+1 } } Hence\quad n=1,a=\frac { 1 }{ 30 } } \right] \)
\(=\frac { 1 }{ 30 } \left( \frac { 1! }{ { \left( \frac { 1 }{ 30 } \right) }^{ 2 } } \right) \)
\(=\frac { 1 }{ 30 } \times \frac { 1 }{ { \left( \frac { 1 }{ 30 } \right) }^{ 2 } } =\frac { 1 }{ \frac { 1 }{ 30 } } =30\)
∴ E(X) = 30 miles (in thousands) or 30,000 miles.

Given demand functionp = 122 - 5x - 2x² and x = 6
When x₀ = 6, p₀ = 122-5(6)-2(6)²
= 122-30-72
= 122-102
P₀ = 20
p₀x₀ = 20 \(\times\) 6 = 120
Consumer's Surplus
CS \(=\int _{ 0 }^{ x }{ f(x) } dx-{ p }_{ 0 }{ x }_{ 0 }\)
\(=\int _{ 0 }^{ 6 }{ (122-5x-2{ x }^{ 2 })dx-120 } \)
\(={ \left[ 122x-\frac { 5{ x }^{ 2 } }{ 2 } -\frac { { 2x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 6 }-120\)
\(=122(6)-5\frac { \left( { 6 }^{ 2 } \right) }{ 2 } -2\frac { \left( { 6 }^{ 3 } \right) }{ 3 } -120\)
\(=732-\frac { 180 }{ 2 } -\frac { 432 }{ 3 } -120\)
= 732-90-144-120
= 732-354
C.S = 378 units

Given,
Marginal cost MC = 6 +10x − 6x²
C = \(\int { MC } dx+k\)
= \(\int { (6+10x-{ 6x }^{ 2 })dx+k } \)
= 6x + 5x² − 2x³ + k (1)
when x = 2, C = 12 (given)
12 = 12 + 20 −16 + k
k = -4
C = 6x + 5x² − 2x³ − 4
Average cost = \(\frac { C }{ x } =\frac { 6x+{ 5x }^{ 2 }-2{ x }^{ 3 }+{ 4 } }{ x } \)
= 6 + 5x − 2x² − \(\frac { 4 }{ x } \)

\(\int { { e }^{ x }\left( { x }^{ 2 }+2x \right) dx } =\int { { e }^{ x }\left[ f(x)+f'(x) \right] } dx\)
= e^x f(x) + c
= e^x x² + c
[Take f(x) = x²
\(\therefore\) f'(x) = 2x]

The order of A is 3 \(\times\) 4.
 \(\therefore \) \(\rho (A)\le 3.\)
Let us transform the matrix A to an echelon form

Matrix A	Elementary Transformation
\(A=\left( \begin{matrix} 1 & 1 & 1 \\ 3 & 4 & 5 \\ 2 & 3 & 4 \end{matrix}\begin{matrix} 1 \\ 2 \\ 0 \end{matrix} \right) \) \(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 1 & 2 \end{matrix}\begin{matrix} 1 \\ -1 \\ -2 \end{matrix} \right) \) \(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & -1 & -2 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} 1 \\ 1 \\ -1 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }-{ 3R }_{ 1 }\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ 2R }_{ 1 }\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }\)

The number of non zero rows is 3.
\(\therefore \) \(\rho (A)=3.\) 

The highest derivative is third order and its power is one
∴ order : 3,
degree : 1